Please help? If Sin A = 3/4 Then find cosA and tanA. Thanks?

x² = 4² - 3²
x² = 7
x = √7

cos A = √7 / 4 , tan A = 3 / √7

sin A = 3/4
cos A = √7/4
tan A = 3/√7

sinA = (3/4). Now sin is pos in quads 1 & 2. We first consider A in quad 1. Then both cos & tan are pos
with cosA = [1 - (3/4)^2]^(1/2) = (1/4)rt7 and tanA = sinA/cosA = (3/4)/[(1/4)rt7] = (3/7)rt7. We secondly
consider A in quad 2 where cos & tan are both neg & = [-(1/4)rt7 , -(3/7)rt7] respectively.

if sin(A) = 3/4, then we know the 4 can be a hypotenuse of a right triangle with the 3 being the opposite. So we can use Pythagorean Theorem to find the adjacent side:

a² + b² = c²
a² + 3² = 4²
a² + 9 = 16
a² = 7
a = √7

So now that we know that, we can find cos(A) and tan(A):

cos() = adj/hyp and tan() = opp/adj
cos(A) = √(7) / 4 and tan(A) = 3/√7

simplifying:

cos(A) = √(7) / 4 and tan(A) = (3/7)√7

As a sanity test, let's see what "A" would be:

A ≈ 48.59°

so:

cos(48.59) = 0.6614 and tan(48.59) = 1.134

Testing:

√(7) / 4 = 0.6614 and (3/7)√7 = 1.134

So I'm correct.

Very straightforward.

If sinA = 3/4, then the ratio of opp/hyp = 3/4.

Let's just say they are
opp = 3 and
hyp = 4.

If you imagine that as an angle in a right triangle, the adj would be √4^2 - 3^2 = √16 - 9 = √7
adj = √

So cosA = adj/hyp √7 / 4

And tan = opp/adj = 3/√7 = (3√7)/7

SinA = opp. / hyp.

Find your third side = a^2 + b^2 = c^2

3^2 + b^2 = 4^2
9 + b^2 = 16
b^2 = 7
b = sqrt(7)

cosA = adj. / hyp.
cosA = sqrt(7) / 4

tanA = opp. / adj.
tanA = 3 / sqrt(7)

find the Sin ^-1 of 3/4 and you have A, then just plug A into the other equations

cos(A) = √(7)/4
tan(A) = 3/√(7)

Is A in quadrant I or in quadrant II?
You will get different answers.

This is killing me. Some fairly heavy hitters are responding, but no one is even considering the possibility of negative numbers. This shows a limitation of the right-triangle model. Not all angles are acute, and the argument of a trigonometric function is not necessarily an angle anyway.

cos²(A) + sin²(A) = 1
cos²(A) + (3/4)² = 1
cos²(A) + 9/16 = 1
cos²(A) = 7/16
cos(A) = ±√(7)/4

tan(A) = sin(A)/cos(A)
tan(A) = (3/4) / [±√(7)/4]
tan(A) = 3 / [±√(7)]
tan(A) = ±3√(7)/7

We cannot know the signs of cos(A) or tan(A), but they must have the same sign.

cos(A) = ±√(7)/4
tan(A) = ±3√(7)/7